a) Xét tam giác ABC có:
\[\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180{}^\circ \Rightarrow 2\widehat{ACB}+\widehat{ACB}+90{}^\circ =180{}^\circ \Rightarrow 3\widehat{ACB}=90{}^\circ \Rightarrow \widehat{ACB}=30{}^\circ \]
Ta có: $\widehat{ABC}=2.\widehat{ACB}=2.30{}^\circ =60{}^\circ $
b) Ta có
$\left. \begin{align}
& HK\bot AB \\
& AC\bot AB \\
\end{align} \right\}\Rightarrow HK//AC$
Xét tam giác AHC có:
$\widehat{AHC}+\widehat{HAC}+\widehat{HCA}=180{}^\circ \Rightarrow 90{}^\circ +\widehat{HAC}+\widehat{HCA}=180{}^\circ \Rightarrow \widehat{HAC}+\widehat{HCA}=90{}^\circ $ (1)
Xét tam giác ABC có:
\[\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180{}^\circ \Rightarrow \widehat{ABC}+\widehat{ACB}+90{}^\circ =180{}^\circ \Rightarrow \widehat{ABC}+\widehat{ACB}=90{}^\circ \] (2)
Từ (1) và (2) suy ra \[\widehat{HAC}=\widehat{ABC}\]
Vì HK // AC nên \[\widehat{HAC}=\widehat{AHK}\](2 góc so le trong)
Vậy \[\widehat{HAC}=\widehat{ABC}=\widehat{AHK}\]